# Spectroscopic Determination of an Equilibrium Constant [closed]

I got values of $$\pu{5.0E-4}$$, $$\pu{2.5E-4}$$, and $$183$$ for the following problems. However, these seem to be inaccurate and I'm not sure what to do.

Here is the procedure:

Obtain 6 disposable test tubes. Label the test tubes 1 through 6. To each of the $$\pu{6 mm}$$ test tubes, add $$\pu{5.0 mL}$$ of $$\pu{2.0E–4 M}$$ $$\ce{KSCN}$$. If you make a mistake, you must rinse out and thoroughly dry the test tube before starting over.

To test tube 1, add $$\pu{5.0 mL}$$ of $$\pu{0.20 M}\; \ce{Fe(NO3)3}$$ and stir. The resulting solution is your standard solution of $$\ce{FeSCN^2+}$$.

For the other test tubes, do the following: Measure $$\pu{10.0 mL}$$ of $$\pu{0.20 M}$$ $$\ce{Fe(NO3)3}$$ in a $$\pu{25 mL}$$ graduated cylinder, fill to $$\pu{25.0 mL}$$ with distilled water, and stir thoroughly to mix. Measure $$\pu{5.0 mL}$$ of the resulting $$\pu{0.080 M}$$ $$\ce{Fe^3+}$$ into test tube 2.

Discard all but $$\pu{10.0 mL}$$ of the $$\pu{0.080 M}$$ $$\ce{Fe^3+}$$, refill the graduated cylinder with $$\pu{10.0 mL}$$ of the $$\pu{0.080 M}$$ $$\ce{Fe^3+}$$ solution to $$\pu{25.0 mL}$$ with distilled water, and stir thoroughly. Add $$\pu{5.0 mL}$$ of the resulting $$\pu{0.032 M}$$ $$\ce{Fe^3+}$$ to test tube 3.

Again discard all but $$\pu{10.0 mL}$$ of the contents of the graduated cylinder, refill to $$\pu{25.0 mL}$$ with distilled water, stir, and add $$\pu{5.0 mL}$$ to test tube 4.

Repeat above procedure with test tubes 5 and 6.

1. Calculate the initial concentrations of $$\ce{Fe^3+}$$ and $$\ce{SCN-}$$ for the six solutions to be prepared in this experiment (i.e., the 6 solutions prepared in test tubes 1–6). These are the concentrations just after the two solutions are mixed in the large test tube (thereby diluting each other) but before any $$\ce{FeSCN^2+}$$ has formed. Assume that the volumes are additive and that the $$\ce{Fe(NO3)3}$$ and $$\ce{KSCN}$$ are each completely dissociated. What is the concentration of $$\ce{Fe^3+}$$ in test tube #5?

$$\begin{array}{c|cc} \hline \text{Test tube} & \text{Initial}~[\ce{Fe^3+}] & \text{Initial}~[\ce{SCN-}] \\ \hline 1 & & \\ 2 & & \\ 3 & & \\ 4 & & \\ 5 & & \\ 6 & & \\ \hline \end{array}$$

1. For the chart in question 1, what is the molar concentration of $$\ce{SCN-}$$ in tube #4?

2. If the equilibrium $$\ce{[FeSCN^2+]}$$ in test tube 2 was determined to be $$\pu{8.3E-5 M}$$, calculate the equilibrium $$\ce{[Fe^3+]}$$.

You must first calculate dilution factors. Assuming the volumes are additive, we can calculate dilution using $$M_1V_1=M_2V_2$$ equation. Also, assuming $$\ce{Fe(NO3)3}$$ and $$\ce{KSCN}$$ are each completely dissociated, we can say:

$$\ce{[Fe(NO3)3] = [Fe^3+]} \text{ and } \ce{[KSCN] = [SCN-]}$$

Thus, initial $$\ce{[Fe^3+]}$$ and $$\ce{[SCN-]}$$ are $$0.20$$ and $$\pu{2.0 \times 10^{-4} M}$$, respectively.

In test tube one: $$\pu{5.0 mL}$$ of $$\pu{0.20 M} \; \ce{Fe(NO3)3}$$ and $$\pu{5.0 mL}$$ of $$\pu{2.0 × 10^{−4} M} \; \ce{KSCN}$$ are added together, thus dilution factor is 1:2. Therefore, using $$M_1V_1=M_2V_2$$ equation you can calculate initial concentrations of $$\ce{[Fe^3+]}$$ and $$\ce{[SCN-]}$$ after the addition (but before they react with each other) as follows:

$$\ce{[Fe^3+]} = \pu{\frac{0.2 \times 5.0}{10.0} M} = \pu{0.10 M} \text{ and } \ce{[SCN-]} = \pu{\frac{2.0 × 10^{−4} \times 5.0}{10.0} M} = \pu{1.0 × 10^{−4} M}$$

As a matter of fact, since the original solution of $$\ce{KSCN}$$ did not undergo any dilution maneuver before final 1:1 dilution, $$\ce{[SCN-]}$$ of all 6 test tubes are same, and its value is $$\pu{1.0 × 10^{−4} M}$$.

On the other hand, the original solution of $$\ce{Fe(NO3)3}$$ has undergone serial dilution of $$\pu{10 mL}$$ to $$\pu{25 mL}$$ (hence dilution factor is $$10/25$$ or $$2/5$$). Therefore, if original concentration of each solution is $$x$$ before add together in test tubes, the final $$\ce{[Fe^3+]}_a$$ before add with $$\ce{[SCN-]}$$ is:

$$\ce{[Fe^3+]}_a = \frac{x \times 2.0}{5.0} \: \pu{M}$$

Thus, the final $$\ce{[Fe^3+]}_a$$ before add with $$\ce{[SCN-]}$$:

$$\ce{[Fe^3+]}_2 = \pu{\frac{0.20 \times 2.0}{5.0} M}= \pu{0.08 M}, \, \ce{[Fe^3+]}_3 = \pu{\frac{0.08 \times 2.0}{5.0} M}= \pu{0.032 M}, \, \ce{[Fe^3+]}_4 = \pu{\frac{0.032 \times 2.0}{5.0} M}= \pu{0.0128 M}, \, \ce{[Fe^3+]}_5 = \pu{\frac{0.0128 \times 2.0}{5.0} M}= \pu{0.00512 M}, \, \ce{[Fe^3+]}_6 = \pu{\frac{0.00512 \times 2.0}{5.0} M}= \pu{0.002048 M}$$

Finally, those concentrations were further diluted in 1:1 ratio in test tubes, and therefore, initial $$\ce{[Fe^3+]}$$ are half of those $$\ce{[Fe^3+]}_a$$ values. I tabulated all of them accordingly in following table, as answer to your question 1:

$$\begin{array}{c|cc} \text{Test Tube #} &\text{Initial \ce{[Fe^3+]}} &\text{Initial \ce{[SCN-]}} \\\hline 1 & \pu{0.10 M} & \pu{1.0 × 10^{−4} M}\\ 2 & \pu{0.04 M} & \pu{1.0 × 10^{−4} M}\\ 3 & \pu{0.016 M} & \pu{1.0 × 10^{−4} M}\\ 4 & \pu{0.0064 M} & \pu{1.0 × 10^{−4} M}\\ 5 & \pu{0.00256 M} & \pu{1.0 × 10^{−4} M}\\ 6 & \pu{0.001024 M} & \pu{1.0 × 10^{−4} M}\\\hline \end{array}$$

1. There are 7 different values in the table. But you have given one relevant answer, and it is also incorrect.
2. Find the answer for $$\ce{[SCN-]}$$ in test tube 4 from the table, which is identical in all 6 test tubes.
3. The concern equilibrium is: $$\ce{Fe^3+ + SCN- <=> FeSCN^2+}$$ Reaction mole ratio of $$\ce{[Fe^3+]}$$ : $$\ce{[SCN-]}$$ : $$\ce{[FeSCN^2+]}$$ is 1 : 1 : 1. If equilibrium concentration of $$\ce{[FeSCN^2+]}$$ in test tube 2 is $$\ce{ 8.3×10^{−5} M}$$, the equilibrium concentration of $$\ce{[Fe^3+]}$$ in test tube 2 is $$\ce{ (0.04 - 8.3×10^{−5}) M}$$ and that of $$\ce{[SCN-]}$$ in test tube 2 is $$\ce{ (1.0×10^{−4} - 8.3×10^{−5}) M}$$. Now, you may able to calculate equilibrium constent for the reaction using following equation:

$$K_\mathrm{eq} = \frac{\ce{[FeSCN^2+]}}{\ce{[Fe^3+]}\ce{[SCN-]}}$$

• @Denise If you find the answer useful, the best way to appreciate the effort that went into producing one would be to upvote and accept the answer. – andselisk Oct 22 '19 at 5:57